Search found 665 matches
- Tue Mar 26, 2013 11:52 pm
- Forum: Geometry
- Topic: One of the Coolest Problems I've Ever Seen (though many may
- Replies: 7
- Views: 5927
Re: One of the Coolest Problems I've Ever Seen (though many
Let $D'$ be the diametrically opposite point point of $D$. Hence $FX$ goes through $D'$. Let $D'D' \cap EF=R'$. Applying Pascal's theorem on cyclic hexagon $FFEDD'D'$ we get $S,R',T$ are collinear. So $ST,D'D',EF$ are concurrent at $R'$ i.e. $R'=ST \cap EF$ which implies $R=R'$. So $\angle RD'I=90^{...
- Thu Mar 21, 2013 10:19 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO-2008-6
- Replies: 1
- Views: 2913
Re: IMO-2008-6
Solution (with proper incentives!): Lemma: Let the incircle of $\triangle ABC$ touch $BC$ at $D$. Let the diametrically opposite point of $D$ be $D'$. $AD' \cap BC=E$. Then $BD=CE$.$E$ is the touch point of the $A$-excircle with $BC$. The converse is also true. Let the touch points of $\omega$ with ...
- Thu Mar 21, 2013 9:48 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO-2008-6
- Replies: 1
- Views: 2913
IMO-2008-6
Let $ABCD$ be a convex quadrilateral with $AB \neq BC$. Denote by $\omega_1$ and $\omega_2$ the incircles of triangles $ABC$ and $ADC$. Suppose that there exists a circle $\omega$ inscribed in angle $ABC$, tangent to the extensions of line segments $AD$ and $CD$. Prove that the common external tange...
- Thu Mar 07, 2013 7:36 pm
- Forum: Social Lounge
- Topic: APMO
- Replies: 2
- Views: 4454
Re: APMO
https://www.facebook.com/photo.php?fbid=10200811850637435 I think this facebook post+ advertise should explain enough. If there any particular detail you couldn't understand, please do post it here. অবস্থা দেখে মনে হচ্ছে না যে এপিএমওর আগে ক্যাম্প শুরু হবে। না হলে ক্যাম্পারদের কি প্রথম আলো অফিসে গিয...
- Thu Mar 07, 2013 12:14 pm
- Forum: Number Theory
- Topic: Easy and Nice
- Replies: 1
- Views: 2286
Re: Easy and Nice
$m(4m^2+m+12)=3(p^n-1) \Rightarrow (m^2+3)(4m+1)=3p^n$. Note that $m=1$ is not a solution and $m^2+3>4m+1 \forall m>1$. If $G.C.D.(m^2+3,4m+1)=1$, then $m^2+3=p^n,4m+1=3$ which doesn't provide any solution. Now let $G.C.D.(m^2+3,4m+1)=d>1$. $d|4(m^2+3)-m(4m+1)=12-m \Rightarrow d|4(12-m)+(4m+1)=49$. ...
- Sun Mar 03, 2013 4:10 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 114874
Re: IMO Marathon
Problem 31:A function $f:\mathbb N\to \mathbb N$ satisfies the following conditions: 1. For any two integers $a,b$ with $\gcd(a,b)=1$, \[f(ab)=f(a)f(b)\] 2. For any two primes $p,q$, \[f(p+q)=f(p)+f(q)\] Prove that, $f(3)=3\;$ and $f(1999)=1999.$ Source: France TST 2000 Let $P(x,y) \Rightarrow f(xy...
- Sun Mar 03, 2013 4:07 pm
- Forum: Geometry
- Topic: A Very Nice Problem
- Replies: 11
- Views: 8031
Re: A Very Nice Problem
Actually Mahi's avatar also symbolizes another popular anime character. Do you remember Gaara especially his forehead seal ?FahimFerdous wrote:My avatar is the sign of an anime character. He's awesome!
- Wed Feb 27, 2013 1:00 pm
- Forum: Geometry
- Topic: Prove concyclic
- Replies: 4
- Views: 4218
Re: Prove concyclic
My solution is pretty computational. At first I would like to note that in my figure $CD \cap BA=H$. From the property of harmonic division we get $\frac{DH}{CH}=\frac{DG}{CG}=\frac{DM-GM}{DM+GM}$ $\Rightarrow DH.DM+DH.GM=CH.DM-CH.GM \Rightarrow DM.DM+DM.GM=(DH+2DM)DM-(DH+2DM)GM$ $\Rightarrow DM^2=G...
- Wed Feb 27, 2013 12:50 pm
- Forum: Algebra
- Topic: Uzbekistan TST 2012-PROBLEM-2
- Replies: 7
- Views: 4806
Re: Uzbekistan TST 2012-PROBLEM-2
Straight-forward Lagrange's Interpolation! Hmm, impressive.zadid xcalibured wrote:$f(x)=f(a) \frac{(x-b)(x-c)}{(a-b)(a-c)}+f(b) \frac{(x-a)(x-c)}{(b-a)(b-c)}+f(c) \frac{(x-a)(x-b)}{(c-a)(c-b)}$
Now plugging $x=a+b+c$ we get $f(a+b+c)$ .
- Tue Feb 26, 2013 10:28 pm
- Forum: Geometry
- Topic: Prove concyclic
- Replies: 4
- Views: 4218
Re: Prove concyclic
I don't think so.Nadim Ul Abrar wrote:Let $AB \cap DC =H$
Note that $D,G,C,H$ build a harmonic range .
So
$\displaystyle \frac{DM}{CM}=\frac{DH}{CH}$